Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $a = \dfrac{n - 6}{-3n^2 - 9n + 162} \times \dfrac{n^3 + 16n^2 + 63n}{n^2 - 8n} $
Answer: First factor out any common factors. $a = \dfrac{n - 6}{-3(n^2 + 3n - 54)} \times \dfrac{n(n^2 + 16n + 63)}{n(n - 8)} $ Then factor the quadratic expressions. $a = \dfrac {n - 6} {-3(n + 9)(n - 6)} \times \dfrac {n(n + 9)(n + 7)} {n(n - 8)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {(n - 6) \times n(n + 9)(n + 7) } { -3(n + 9)(n - 6) \times n(n - 8)} $ $a = \dfrac {n(n + 9)(n + 7)(n - 6)} {-3n(n + 9)(n - 6)(n - 8)} $ Notice that $(n + 9)$ and $(n - 6)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {n\cancel{(n + 9)}(n + 7)(n - 6)} {-3n\cancel{(n + 9)}(n - 6)(n - 8)} $ We are dividing by $n + 9$ , so $n + 9 \neq 0$ Therefore, $n \neq -9$ $a = \dfrac {n\cancel{(n + 9)}(n + 7)\cancel{(n - 6)}} {-3n\cancel{(n + 9)}\cancel{(n - 6)}(n - 8)} $ We are dividing by $n - 6$ , so $n - 6 \neq 0$ Therefore, $n \neq 6$ $a = \dfrac {n(n + 7)} {-3n(n - 8)} $ $ a = \dfrac{-(n + 7)}{3(n - 8)}; n \neq -9; n \neq 6 $